# compute limit (no l'Hospital rule)

I need to compute $$\lim_{x\to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2x}.$$ I can not use the l'Hospital rule.

• Your sin(x)^2 is sin(x^2) or $\sin^2x$? – mathlove Jan 5 '14 at 16:35
• Can you use Taylor series? – Gerry Myerson Jan 5 '14 at 16:36

As lcm $(2,3)=6,$ let $(\cos x)^\frac16=y$

$\displaystyle\implies (\cos x)^\frac12=y^3,(\cos x)^\frac13=y^2$ and $\displaystyle \cos x=y^6\implies\sin^2x=1-y^{12}$

$$\lim_{x\to0}\frac{(\cos x)^\frac12-(\cos x)^\frac13}{\sin^2x}=\lim_{y\to1}\frac{y^3-y^2}{1-y^{12}}=\lim_{y\to1}\frac{-y^2(1-y)}{(1-y)(1+y+\cdots+y^{10}+y^{11})}$$

We can cancel $1-y$ as $1-y\ne0$ as $y\to1$

Alternatively, $$\lim_{y\to1}\frac{y^3-y^2}{1-y^{12}}=\lim_{y\to1}(-y^2)\cdot\frac1{\lim_{y\to1}\frac{y^{12}-1}{y-1}}$$

Again, $$\lim_{y\to1}\frac{y^{12}-1}{y-1}=\lim_{y\to1}\frac{y^{12}-1^{12}}{y-1}=\frac{d(y^{12})}{dy}_{(\text{ at }y=1)}$$

• Quick question: Why in the original function when we've got $0/0$ we had to go on with factoring, but when we came to the limit of $\frac{y-1}{y-1}$ times $-1/12$ (the result of the other limit), we could say that the $-1/12$ is the end result, disregarding that $\frac{y-1}{y-1}$ in the limit is still $0/0$? – SasQ Jan 6 '14 at 16:07
• @SasQ, $y\to1$ is not synonymous of $y=1$ there is infinitesimal but non-zero difference between $y$ and $1$. See mathsisfun.com/calculus/limits.html and math.toronto.edu/lgoldmak/A31F12/summary.html – lab bhattacharjee Jan 6 '14 at 18:30

$$\lim_{x\to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2x}=\lim_{x\to 0} \frac{\sqrt{\cos x} -1+1- \sqrt[3]{\cos x}}{\sin^2x}=\lim_{x\to 0} \frac{\sqrt{1+\cos x-1} -1- (\sqrt[3]{1+\cos x-1}-1)}{\cos x-1}\cdot\frac{\cos x-1}{\sin^2x}=\lim_{x\to 0}[\frac{(1+\cos x-1)^\frac{1}{2}-1}{\cos x-1}-\frac{(1+\cos x-1)^\frac{1}{3}-1}{\cos x-1}]\cdot\frac{-1}{2}=(\frac{1}{2}-\frac{1}{3})\cdot\frac{-1}{2}=-\frac{1}{12}$$

We applied:$$\lim_{t\to 0}\frac{(1+t)^r-1}{t} =r$$